Problem 38 - Precalculus: Mathematics for Calculus

Question

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example 6. $$\left\\{\begin{array}{r} -3 x+5 y=2 \\ 9 x-15 y=6 \end{array}\right.$$

Step-by-Step Solution

Answer

The system has infinitely many solutions. Expressed as \( \left( \frac{5t - 2}{3}, t \right) \).

Step 1: Write the System of Equations

Write down the system of equations:1. \(-3x + 5y = 2\) 2. \(9x - 15y = 6\)

Step 2: Multiply First Equation

To analyze the system, multiply the first equation by 3 to compare with the second equation:\[ 3(-3x + 5y) = 3 \times 2 \]This gives us:\[-9x + 15y = 6\]

Step 3: Compare Equations

Now compare the transformed first equation \(-9x + 15y = 6\) with the second equation \(9x - 15y = 6\). You will observe both equations are identical.

Step 4: Analyze for Solutions

Since both equations are identical after transformation, the system is dependent and has infinitely many solutions.

Step 5: Express Infinitely Many Solutions

Express the solutions in terms of one variable, say \(y\). Let \(y = t\) (where \(t\) is any real number).Substitute \(y = t\) into the original first equation: \[-3x + 5t = 2\]Solve for \(x\): \[x = \frac{5t - 2}{3}\]Thus, the solutions can be represented as ordered pairs: \[\left( \frac{5t - 2}{3}, t \right)\]

Key Concepts

Dependent System

When we talk about a dependent system of linear equations, we are referring to a situation where the equations are not independent of each other. In simpler terms, one equation is a multiple or a rearrangement of another, meaning that they essentially describe the same line on a graph. In such cases, every solution of one equation is a solution of the other.

A dependent system arises in our exercise example when we transform the first equation by multiplying it. After doing this, we find that both equations become identical. This tells us that instead of having two separate equations, we actually have just one equation expressing the same relationship between the variables. Consequently, this overlap means the system is dependent. Thus, the system of equations does not lead to a unique solution but rather a set of solutions that satisfy the same line.

Infinitely Many Solutions

A key characteristic of a dependent system is that it does not have a single, unique solution. Instead, it has infinitely many solutions. What does this mean? In the context of equations, infinitely many solutions imply that there are countless pairs of numbers that can satisfy both equations simultaneously.

In our exercise, we observe that after adjusting the equations, they turn out to be identical. This indicates that any point (x, y) on the line defined by the equation satisfies both equations. Since there are infinitely many points on a line, we have infinitely many solutions.

• Visualize this as every point on the line described by the equation being a solution.
• This is why expressing solutions in terms of a variable, like setting y = t, helps simplify how we represent this concept.
• Our ordered pairs then show this continuous range of solutions as \((\frac{5t - 2}{3}, t)\).

In this scenario, the solutions form a continuum, an unbroken series of values that satisfy the system.

Ordered Pairs

Ordered pairs are a common way of representing solutions to a system of equations, especially when dealing with two variables such as x and y. An ordered pair is expressed as (x, y), where x is the first element and y is the second.

In our exercise, after realizing the system has infinitely many solutions, we use ordered pairs to present these solutions in a clear format. By setting one variable, such as y equal to a parameter t, we can describe all solutions systematically:

• The y-coordinate (or second element of the pair) is simply the parameter t.
• The x-coordinate is calculated using the given equation. In this case, \(x = \frac{5t - 2}{3}\).

Hence, the solutions can be expressed in the form of ordered pairs such as \((\frac{5t - 2}{3}, t)\), where t is any real number.

This method not only simplifies the presentation of all possible solutions but also illustrates how by choosing any value for t (or y), the corresponding x value can be determined to keep the solution valid within the system.